Equation of tangent to #160; y^2=4ax is #160; y=mx+a/m Let it passes through P(h, k) ∴ m^2h mk+a=0 (ii) #160; tanθ _0=m_1+m_2/1 m_1m_2=k/ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Integration of Trigonometric Functions

ii θ 1+θ 2=...

Question

(ii) $θ_{1} + θ_{2} = θ_{0}$ (const)

Open in App

Solution

Equation of tangent to $y^{2} = 4 a x$ is $y = m x + \frac{a}{m}$
Let it passes through P(h, k)
$∴ m^{2} h - m k + a = 0$
(ii) $tan θ_{0} = \frac{m_{1} + m_{2}}{1 - m_{1} m_{2}} = \frac{k / h}{1 - a / h}$ $\Rightarrow y = (x - a) tan θ_{0}$

Suggest Corrections

Similar questions

Q. If

\frac{{cos}^{4} θ}{θ_{1}} + \frac{{sin}^{4} θ}{θ_{2}} = \frac{1}{θ_{1} + θ_{2}}

,then

\frac{θ_{2} + θ_{1}}{θ_{1}}

equals

Q. If

\frac{{cos}^{4} x}{θ_{1}} + \frac{{sin}^{4} x}{θ_{2}} = \frac{1}{θ_{1} + θ_{2}},

then

\frac{θ_{2}}{θ_{1}}

equals

Q. A simple pendulum is made of a string of length

l

and a bob of mass

m

, is released from a small angle

θ_{0}

. It strikes a block of mass

M

, kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle

θ_{1}

. Then,

M

is given by

If $t a n θ_{1}, t a n θ_{2}, t a n θ_{3}$ are the real roots of $x^{3} - (a + 1) x^{2} + (b - a) x - b = 0 w h e r e θ_{1}, θ_{2}, θ_{3}$ are acute then $θ_{1} + θ_{2} + θ_{3} =$

Q. If θ1,θ2,θ3........,θn are in A.P., whose common difference is d, then,
secθ1secθ2+secθ2secθ3+..........+secθn−1secθn=

Join BYJU'S Learning Program

(ii) θ1+θ2=θ0 (const)

Equation of tangent to y2=4ax is y=mx+amLet it passes through P(h, k)∴m2h−mk+a=0(ii) tanθ0=m1+m21−m1m2=k/h1−a/h ⇒y=(x−a)tanθ0

(ii) $θ_{1} + θ_{2} = θ_{0}$ (const)

Equation of tangent to $y^{2} = 4 a x$ is $y = m x + \frac{a}{m}$
Let it passes through P(h, k)
$∴ m^{2} h - m k + a = 0$
(ii) $tan θ_{0} = \frac{m_{1} + m_{2}}{1 - m_{1} m_{2}} = \frac{k / h}{1 - a / h}$ $\Rightarrow y = (x - a) tan θ_{0}$